Richard Clark wrote:

On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:

At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.


Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

I've gotten some education from it, and had some fun too!

- Mike KB3EIA -

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.
tnx


--
73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:
At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.

Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

73's
Richard Clark, KB7QHC

On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik"
wrote:

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.
tnx

Hi Hank,

There is no answer as you phrase the question because the transmitter
does dissipate power reflected to it. I can easily imagine you may
find no understanding as I go hyperbolic in the following lines. :-)

It can also reflect (some or all of) what it does not absorb, but only
if it does not match the line connected to it. The degree to which it
absorbs/reflects is determined by the ratio of its Z to the line/load
at the antenna terminal. In this case (bear with me), the source Z is
NOW found at the load's mismatch, transformed through the line. This
means matching works both ways (often the singlemost ignored aspect of
obvious reciprocity in any of these arguments). Such circuit analysis
is called superposition. This is merely an academic way of saying you
look at the problem from both points of view where you reverse roles
(source becomes load, and load becomes source). This is a common
practice learned in first quarter circuit analysis when Kirchhoff's
laws are developed (Thevenin/Norton equivalent circuits also emerge).

For example (a strained one at that), if you had a 500 Ohm Source
characteristic Z, you will never launch much power into a 50 Ohm
system because it would immediately hit a reflective interface at the
antenna connector. This means that a 500 Ohm source, when confronted
by power going towards it from a 50 Ohm system will reflect most of
that power (but how did we get this power into the system in the first
place - Karma?) This, by design, will never happen in any transistor
ham rig built in the last quarter century.

So, this is why you see such a vacuum of response when you (the
general readership "you") ask:
"What is the source Z if it is not 50 Ohms?"
Silence guarantees they either don't know (a fatal admission for egos)
or if they offered a value, we could all balance the checkbook and
that would end the game being played. Hence we are treated to all
these suppositions of shorts and opens or magic reflectors hidden
beneath the hood. The frequent toss-off comment of "no one knows" is
called projection, a psychological salve for the ego meaning if I
don't know, then certainly no one else does either - or they are
wrong.

The answer is the transmitter source Z is 50 Ohms at rated power. If
a watt of power is chooglin' down the line toward it, that 50 Ohms is
going to dissipate into a watt worth of calories. This can be argued
with wave mechanics, or lumped circuit equivalents - doesn't matter
because it's all the same calories. Modern rigs can tolerate this
watt through limited feedback and level controlling circuitry - if you
paid more than a kilobuck for your rig that is. For the rest of us,
it's a spin of the wheel and you take your chance. 1 watt hardly
amounts to much, but are you that lucky that it is ONLY 1 watt? Are
we to suppose those unfortunate souls who blasted their rig
transmitting into a mismatch lost it only because they lacked enough
magic pixie dust? I will bet no rig was lost to a cold snap with
frosty finals.

I've already answered about where the heat can be found, so we will
conserve bandwidth.

=== WARNING to the logic impaired, the following is a supposition ===

Now, lets simply accept those answers that require the magic pixie
dust of total reflection from the transmitter. Fine, the mismatched
antenna reflects some power, this power returns to the transmitter,
the transmitter simple routes it ALL back to the antenna,
round-and-round until the antenna finally radiates it. If this were
true, what do we need tuners for? Take a survey of everyone who
chants this mantra of the rig reflecting, and ask if they have a tuner
in the line. Is it a paper weight holding down their license? Is it
a line stretcher because they needed another foot extension between
the antenna lead and their rig? Dare I point out the utter failure of
their faith?

I will pause to allow those answers to emerge and enjoy the thrashing
dawn of a new era of metaphysics. Ooooohhhhhhmmmmms Laaaaaaaaaaw!

73's
Richard Clark, KB7QHC

Henry Kolesnik wrote:
You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.

By definition, reflected energy dissipated in the source was never
generated in the first place.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:
You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So,
would
you be kind enough to give me a better understanding of the "mechanism"
in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that
acts
like a checkvalve or diode that reflects.

By definition, reflected energy dissipated in the source was never
generated in the first place.
--
73, Cecil http://www.qsl.net/w5dxp


That's not a very good definition.
Would you say that a rock, thrown vertically, never was thrown just because
it returned to hit you on the head?

Ed
wb6wsn

Post below

REALITY CHECK: If tx source Z really was 50 ohms, a tx connected to a 50
ohm load would lose 1/2 of the RF power it generates to that internal Z.
And if that is true, then the heat produced by the tx, and the AC power
consumed by the tx would be proof of that.

However careful measurement of heat production and AC power consumption of
broadcast txs (for example) does NOT support the belief that the source Z of
a tx designed for 50 ohm loads IS 50 ohms.

Following is the output of an old DOS program I wrote years ago to show
operating parameters for a 25kW FM broadcast tx manufactured by my employer,
when operating at any RF power and line voltage in its rated range. These
values closely track the average of factory test data accumulated over 100s
of delivered units.

In this example, if the tx had a 50 ohm source Z it would have to generate
50kW of RF in order to deliver 25kW to a 50 ohm load. In the first place,
it doesn't have a big enough power supply or a big enough PA to do that.
But even if it could, the input power and heat production would be almost
twice the real values shown below.

The same physics applies to ham transmitters, which should be verifiable by
an unbiased evaluation of comparable operating parameters.

- RF

* FM TRANSMITTER AC LOAD *

Transmitter name (HT-??FM): ? 25
TPO in kW: ? 25
AC line voltage: ? 220

RESULT:

Total Tx = 103.7 amps/phase, 39.5 kVA total.
H.V. Power Supply Current = 92.7 amps/phase
Total AC/RF Efficiency = 66.5 %

PA Air = 400 cu. feet/min.
Tx Cabinet Flushing Air = 335 cu. feet/min.
Power Supply Air = 250 cu. feet/min.
Total Air = 985 cu. feet/min.

Total Heat = 42810 BTU/Hr.
Heat Rise, total tx system = 39 deg. F.
Air Conditioner Load = 3.6 tons

____________

"Richard Clark"
"What is the source Z if it is not 50 Ohms?"
(clippage)
The answer is the transmitter source Z is 50 Ohms at rated power.

Richard
This is in response to your answer of last night. Before going to bed I got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking was
that the reflected power enters the amplifier, causing tube overheating and
destruction. However, I dispelled this misconception in the above mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".
I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR


"Richard Clark" wrote in message
...
On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik"
wrote:

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So,
would
you be kind enough to give me a better understanding of the "mechanism"
in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that
acts
like a checkvalve or diode that reflects.
tnx

Hi Hank,

There is no answer as you phrase the question because the transmitter
does dissipate power reflected to it. I can easily imagine you may
find no understanding as I go hyperbolic in the following lines. :-)

It can also reflect (some or all of) what it does not absorb, but only
if it does not match the line connected to it. The degree to which it
absorbs/reflects is determined by the ratio of its Z to the line/load
at the antenna terminal. In this case (bear with me), the source Z is
NOW found at the load's mismatch, transformed through the line. This
means matching works both ways (often the singlemost ignored aspect of
obvious reciprocity in any of these arguments). Such circuit analysis
is called superposition. This is merely an academic way of saying you
look at the problem from both points of view where you reverse roles
(source becomes load, and load becomes source). This is a common
practice learned in first quarter circuit analysis when Kirchhoff's
laws are developed (Thevenin/Norton equivalent circuits also emerge).

For example (a strained one at that), if you had a 500 Ohm Source
characteristic Z, you will never launch much power into a 50 Ohm
system because it would immediately hit a reflective interface at the
antenna connector. This means that a 500 Ohm source, when confronted
by power going towards it from a 50 Ohm system will reflect most of
that power (but how did we get this power into the system in the first
place - Karma?) This, by design, will never happen in any transistor
ham rig built in the last quarter century.

So, this is why you see such a vacuum of response when you (the
general readership "you") ask:
"What is the source Z if it is not 50 Ohms?"
Silence guarantees they either don't know (a fatal admission for egos)
or if they offered a value, we could all balance the checkbook and
that would end the game being played. Hence we are treated to all
these suppositions of shorts and opens or magic reflectors hidden
beneath the hood. The frequent toss-off comment of "no one knows" is
called projection, a psychological salve for the ego meaning if I
don't know, then certainly no one else does either - or they are
wrong.

The answer is the transmitter source Z is 50 Ohms at rated power. If
a watt of power is chooglin' down the line toward it, that 50 Ohms is
going to dissipate into a watt worth of calories. This can be argued
with wave mechanics, or lumped circuit equivalents - doesn't matter
because it's all the same calories. Modern rigs can tolerate this
watt through limited feedback and level controlling circuitry - if you
paid more than a kilobuck for your rig that is. For the rest of us,
it's a spin of the wheel and you take your chance. 1 watt hardly
amounts to much, but are you that lucky that it is ONLY 1 watt? Are
we to suppose those unfortunate souls who blasted their rig
transmitting into a mismatch lost it only because they lacked enough
magic pixie dust? I will bet no rig was lost to a cold snap with
frosty finals.

I've already answered about where the heat can be found, so we will
conserve bandwidth.

=== WARNING to the logic impaired, the following is a supposition ===

Now, lets simply accept those answers that require the magic pixie
dust of total reflection from the transmitter. Fine, the mismatched
antenna reflects some power, this power returns to the transmitter,
the transmitter simple routes it ALL back to the antenna,
round-and-round until the antenna finally radiates it. If this were
true, what do we need tuners for? Take a survey of everyone who
chants this mantra of the rig reflecting, and ask if they have a tuner
in the line. Is it a paper weight holding down their license? Is it
a line stretcher because they needed another foot extension between
the antenna lead and their rig? Dare I point out the utter failure of
their faith?

I will pause to allow those answers to emerge and enjoy the thrashing
dawn of a new era of metaphysics. Ooooohhhhhhmmmmms Laaaaaaaaaaw!

73's
Richard Clark, KB7QHC

"Henry Kolesnik" wrote in message
.. .
Richard
This is in response to your answer of last night. Before going to bed I
got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim
from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the
most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking
was
that the reflected power enters the amplifier, causing tube overheating
and
destruction. However, I dispelled this misconception in the above
mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input
of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".
I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR
Henry,
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line, there will be *0 Volts across the resistor*. There will be 0 current
through the resistor, and the reflected wave will be re reflected for all
values of R, including R=Z0, because the reflected wave will not "know" what
R is. You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Tam/WB2TT

Ed Price wrote:
"Cecil Moore" wrote:
By definition, reflected energy dissipated in the source was never
generated in the first place.

That's not a very good definition.
Would you say that a rock, thrown vertically, never was thrown just because
it returned to hit you on the head?

Nope, I wouldn't. But that seems to be what some people
are saying. I objected to that definition about 15 years
ago. I was asked to prove otherwise and was unable to
do so. The generated power is *defined* as the *net*
steady-state output power of the source.

That's why many of my examples involve signal generators
equipped with a circulator and load, outputting a constant
forward voltage in phase with a constant forward current.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Fry wrote:

Post below

REALITY CHECK: If tx source Z really was 50 ohms, a tx connected to a 50
ohm load would lose 1/2 of the RF power it generates to that internal Z.
And if that is true, then the heat produced by the tx, and the AC power
consumed by the tx would be proof of that.

That describes a Class-A amplifier.

However careful measurement of heat production and AC power consumption of
broadcast txs (for example) does NOT support the belief that the source Z of
a tx designed for 50 ohm loads IS 50 ohms.

With other classes of amps, the amplifier's individual component signals
are non-linear.
--
73, Cecil http://www.qsl.net/w5dxp




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